Non-Homogeneous 1-Dimensional Wave Equation with arbitrary initial/boundary conditions 
Solve by direct methods the $1$-dimensional non-homogeneous wave equation
  $$u_{tt} - u_{xx} = f(x,t), \hspace{1cm} u(x,0) = g(x), \hspace{1cm} u_t (x,0) = h(x).$$

Our solution will be of the form $u(x,t) = p(x,t) + o(x,t)$, where $p(x,t)$ is a particular solution satisfying homogeneous boundary conditions:
\begin{align*}
p_{tt} - p_{xx} &= f(x,t) \\
p(x,0) &= 0 \\
p_t (x,0) &= 0 
\end{align*}
and $o(x,t)$ is a solution of the corresponding to the homogeneous PDE:
\begin{align*}
o_{tt} - o_{xx} &= 0 \\
o(x,0) &= g(x) \\
o_t (x,0) &= h(x) 
\end{align*}
By D'Almbert's formula we know that 
$$o(x,t) = \dfrac{1}{2}\left( g(x + t) + g(x - t) + \int_{x-t}^{x+t} h(y)dy\right)$$
So our goal is to find the solution to the particular solution $p$. That is we need to solve
\begin{align*}
p_{tt} - p_{xx} &= f(x,t) \\
p(x,0) &= 0 \\
p_t(x,0) &= 0 
\end{align*}
I know that the solution should be 
$$p(x,t) = \dfrac{1}{2}\int_0^t \int_{x-(t-s)}^{x+(t-s)} f(r,s)drds.$$
where Duhamel's principle should be utilized somewhere.
Question: How do I solve the PDE involving $p$? My lecture notes only have the solution involving $o$ (i.e. derivation of D'Almbert's formula). 
 A: Here is an attempt:
Let $v(x,t) := u_t + u_x$. Then, we have that
$$v_t - v_x = u_{tt} + u_{xt} - u_{xt} - u_{xx} = f(x,t)$$
where
\begin{align*}
v(x,0) &= u_t (x,0) + u_x (x,0) \\
&= h(x) + g'(x)
\end{align*}
Now we let $w(s) = v(x-s,t+s)$, so that 
\begin{align*}
w'(s) &= -v_x(x-s,t+s) + v_t(x-s,t+s) \\
&=f(x-s,t+s)
\end{align*}
Hence,
$$w(\tau) - w(-t) = \int_{-t}^{\tau} f(x-s,t+s)ds$$
so that,
\begin{align*}
w(\tau) &= w(-t) + \int_{-t}^{\tau} f(x-s,t+s)ds \\
&= v(x+t,0) + \int_{-t}^{\tau} f(x-s,t+s)ds
\end{align*} 
Therefore,
$$v(x,t) = w(0) = h(x+t) + g'(x+t) + \int_{-t}^0 f(x-s,t+s)ds$$
So that we're left with:
$$\begin{cases} u_t + u_x &= v(x,t) \\ u_x (x,0) &= g(x) \\ u_t (x,0) &= h(x) \end{cases}$$
Now if we let $z(s):=u(x+s,t+s)$, then:
\begin{align*}
z'(s) &= u_x(x+s,t+s) + u_t(x+s,t+s) \\
&= v(x+s,t+s)
\end{align*}
Therefore,
$$z(\tau) - z(-t) = \int_{-t}^{\tau} v(x+s,t+s)ds$$
\begin{align*}
z(0) = u(x,t) &= u(x-t,0) + \int_{-t}^0 v(x+s,t+s)ds \\
&= g(x-t) + \int_{-t}^0 v(x+s,t+s)ds
\end{align*}
where $v(x,t)=h(x+t) + g'(x+t) + \displaystyle\int_{-t}^0 f(x-s,t+s)ds$. 

Is this correct? 

A: Let $\begin{cases}p=x+t\\q=x-t\end{cases}$ ,
Then $u_x=u_pp_x+u_qq_x=u_p+u_q$
$u_{xx}=(u_p+u_q)_x=(u_p+u_q)_pp_x+(u_p+u_q)_qq_x=u_{pp}+u_{pq}+u_{pq}+u_{qq}=u_{pp}+2u_{pq}+u_{qq}$
$u_t=u_pp_t+u_qq_t=u_p-u_q$
$u_{tt}=(u_p-u_q)_t=(u_p-u_q)_pp_t+(u_p-u_q)_qq_t=u_{pp}-u_{pq}-u_{pq}+u_{qq}=u_{pp}-2u_{pq}+u_{qq}$
$\therefore u_{pp}-2u_{pq}+u_{qq}-(u_{pp}+2u_{pq}+u_{qq})=f\left(\dfrac{p+q}{2},\dfrac{p-q}{2}\right)$
$-4u_{pq}=f\left(\dfrac{p+q}{2},\dfrac{p-q}{2}\right)$
$u_{pq}=-\dfrac{1}{4}f\left(\dfrac{p+q}{2},\dfrac{p-q}{2}\right)$
$u(p,q)=F(p)+G(q)-\dfrac{1}{4}\int_0^q\int_0^pf\left(\dfrac{r+s}{2},\dfrac{r-s}{2}\right)~dr~ds$
$u(x,t)=F(x+t)+G(x-t)-\dfrac{1}{4}\int_0^{x-t}\int_0^{x+t}f\left(\dfrac{r+s}{2},\dfrac{r-s}{2}\right)~dr~ds$
